A body is first heated to 100°C and then kept into a room for cooling. The temperature of the room is 25°C. The time taken to cool the body from 100°C to 70°C is t₁ and then to cool from 70°C to 40°C is t₂. Find the ratio of t₁ to t₂.

CONCEPT:

• Heat always flows from a hot to a cold body.
• A body at a temperature higher than its surrounding cools down by giving away the heat to the surroundings and a body at a temperature lower than its surrounding warms up by taking heat from the surrounding.
• The rate at which a body loses heat to the surrounding is given by Newton's law of cooling.

Newton’s law of cooling:

• According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body, and its surroundings.
• Mathematically it is given as,

\(\Rightarrow \frac{T_1-T_2}{t} = k\left [ \frac{T_1+T_2}{2} - T_s \right ]\)

Where T₁ = temperature before cooling, T₂ = temperature after cooling, T_s = temperature of surrounding, t = time required and 

k = Positive constant that depends on the area and nature of the surface of the body under consideration

CALCULATION:

For case 1: (T₁ = 100 °C, T₂ = 70 oC, T_s = 25 oC)

\(\Rightarrow \frac{100-70}{t_1}=k[\frac{100+70}{2} - 25] \\ \Rightarrow\frac{30}{t_1}=k[85 - 25]\)

\(\Rightarrow \frac{30}{t_1}=60k\)      ----(1)

For case 2: (T₁ = 70 °C, T₂ = 40 °C, T_s = 25 °C)

• The time taken to cool from 70°C to 40°C is

\(\Rightarrow\frac{70-40}{t_2}=k[\frac{70+40}{2} - 25] \\\Rightarrow \frac{30}{t_2}=k[55 - 25]\)

\(\Rightarrow \frac{30}{t_2}=30k\)       ----(2)

By equation 1 and equation 2,

\(\Rightarrow \frac{t_1}{t_2}=\frac{30}{60}\)

\(\Rightarrow \frac{t_1}{t_2}=\frac{1}{2}\)

Hence, option 2 is correct.